Question

$1$ mol of an ideal gas expands reversibly and isothermally from $1\mathrm{l}$ to $100\mathrm{l}$ at $300\mathrm{K}$. The heat exchange during the process is near?

Answer 1

Step 1: Given data

Number of moles of gas = $1$ mole

Temperature = $300\mathrm{K}$

Final volume of gas = $100\mathrm{L}$

Initial volume of gas = $1\mathrm{L}$

Step 2: Formula used

Work done in an isothermal irreversible expansion is given by the formula: $W=-nRT\ln \left(\frac{V_f}{V_i}\right)$

where,

$\mathrm{W}$ = workdone

$\mathrm{R}$= gas constant = 8.314

$\mathrm{T}$= temperature

$\mathrm{n}$ = no. of moles

${\mathrm{V}}_{\mathrm{f}}$ = Final volume of gas

${\mathrm{V}}_{\mathrm{i}}$ = Initial volume of gas

The first law of thermodynamics: $\Delta U=\Delta H+W$ where,

$\Delta U$ = change in internal energy

$\Delta H$ = change in heat energy

$W$ = work done

Step 3: Calculating the work done

$$\begin{gathered} W=-1\times 8.314\times 300\ln \left(\frac{100}{1}\right) \\ =-1\times 8.314\times 300\ln \left({10}^2\right) \\ =-1\times 8.314\times 300\times 2\ln \left(10\right) \\ =-11486.2154779J \\ =-11.48kJ \end{gathered}$$

Step 4: Calculating the heat exchanged

Since the process is isothermal, hence temperature is constant, $\Delta U=0.$

Hence the heat exchange for the process can be calculated as:

$$\begin{gathered} \Rightarrow 0=∆H+W \\ \Rightarrow ∆H=-W \\ =-(-11.48) \\ =11.48kJ \end{gathered}$$

Hence the heat exchanged in the process is 11.48kJ.